Áp dụng bất đẳng thức Cauchy-Schwarz ta có:
\(B=\dfrac{1}{2-x}+\dfrac{1}{x}\ge\dfrac{\left(1+1\right)^2}{2-x+x}=\dfrac{4}{2}=2\)
Dấu "=" xảy ra khi: \(x=1\)
p/s Mình nghĩ đề phải là \(0< x\le1\) nhé
áp dụng bunhia
\(\left[\left(\sqrt{\dfrac{2}{1-x}}\right)^2+\left(\sqrt{\dfrac{1}{x}}\right)^2\right]\left[\left(\sqrt{1-x}\right)^2+\left(\sqrt{x}\right)^2\right]\)
\(\ge\left(\sqrt{\dfrac{2}{1-x}}.\sqrt{1-x}+\sqrt{\dfrac{1}{x}}.\sqrt{x}\right)^2\)
\(\Leftrightarrow\left(\dfrac{2}{1-x}+\dfrac{1}{x}\right)\left(1\right)\ge\left(\sqrt{2}+\sqrt{1}\right)^2\)
\(\Rightarrow B\ge\left(\sqrt{2}+1\right)^2\)
dấu = xảy ra khi \(\dfrac{\dfrac{2}{1-x}}{1-x}=\dfrac{\dfrac{1}{x}}{x}\Leftrightarrow x=\sqrt{2-1}\)
